What is the volume of the solid generated when the area of the curve y=3-3^x is rotated about the x axis

Any ordinates given or assuming first quadrant?

Use the powerful Blue Eyes White logs pls don't kill me

What is the volume of the solid generated when the area of the curve y=3-3^x is rotated about the x axis
Infinity.

If bounds are given, use the formula

$V = \pi \int^a_b y^2\, dx$

and use

$k = e^{\log k}$ (for all $k > 0$)

Well to summ wat the above posts states is the

x = e ^ (lnx)

is the same thing right?

didnt finish it fully, should be right if i didnt make silly errors, but you want to reach there

Herowise one of your coefficients is not correct.

Originally Posted by HeroWise
Well to summ wat the above posts states is the

x = e ^ (lnx)

is the same thing right?

didnt finish it fully, should be right if i didnt make silly errors, but you want to reach there
that is some funky integration going on there:

Might be easier to use change of base before integrating. Also maybe leave the exponent as x rather than x+1 before integrating. Less likely to make mistakes that way.
I'll do an improper integral here:
$\displaystyle\int{9-6 e^{x \ln 3} + e^{x \cdot 2\ln 3}$

$= 9x - \frac{6}{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3} + C$

$= 9x - \frac{2}{\ln 3} 3^{x+1} + \frac{1}{2 \ln 3} 3^{2x}+C$

or

$\displaystyle\int{9-2 e^{(x+1) \ln 3} + e^{x \cdot 2\ln 3}$

$=\displaystyle\int{9-2 e^{\ln 3} \cdot e^{x \ln 3} + e^{x \cdot 2\ln 3} + C$

$= 9x - \frac{2}{\ln 3} e^{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3} + C$

$since e^{\ln 3} = 3$

$= 9x - \frac{6}{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3}+C$

$= 9x - \frac{2}{\ln 3} 3^{x+1} + \frac{1}{2 \ln 3} 3^{2x}+C$

This is the improper integral. The integral with limits is done in a similar fashion.

Pfft i feel dumb, i see where i went wrong, thanks a lot anyway @dan964

EDIT: Wait, no i got the same result as yours @dan964, but i didnt simplify it. if i had, it would come down to your answer, and the ordinates, as i said, im geussing the question is asking for first quadrant, i remember a similar question while going over the Cambridge book, so i just "geussed". That isnt really good but i think the working should suffice the OP

Originally Posted by dan964
Herowise one of your coefficients is not correct.

that is some funky integration going on there:

Might be easier to use change of base before integrating. Also maybe leave the exponent as x rather than x+1 before integrating. Less likely to make mistakes that way.
I'll do an improper integral here:
$\displaystyle\int{9-6 e^{x \ln 3} + e^{x \cdot 2\ln 3}$

$= 9x - \frac{6}{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3} + C$

$= 9x - \frac{2}{\ln 3} 3^{x+1} + \frac{1}{2 \ln 3} 3^{2x}+C$

or

$\displaystyle\int{9-2 e^{(x+1) \ln 3} + e^{x \cdot 2\ln 3}$

$=\displaystyle\int{9-2 e^{\ln 3} \cdot e^{x \ln 3} + e^{x \cdot 2\ln 3} + C$

$= 9x - \frac{2}{\ln 3} e^{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3} + C$

$since e^{\ln 3} = 3$

$= 9x - \frac{6}{\ln 3} e^{x \ln 3} + \frac{1}{2 \ln 3} e^{2x \ln 3}+C$

$= 9x - \frac{2}{\ln 3} 3^{x+1} + \frac{1}{2 \ln 3} 3^{2x}+C$

This is the improper integral. The integral with limits is done in a similar fashion.
This is what I got you will see the question from the textbook and my solutions I can't seem to get 13c. The books solution is pi(9-8/log3) u^3
https://imgur.com/gallery/u5PCA

Can someone please write the full solution to 13c

Can you tell us the ordinates then, because as the working out provided by both @dan964 and i are elligible, so just put in the ordinates like normal integration, or provide us with ordinates; thatd be helpful for us ty

or send a pic of the whole excercise

$V = \pi \int^1_0 (3-3^x)^2 \, dx$

$= \pi \int^1_0 9 - 6(3^x) + 3^{2x}\, dx$

$= \pi \int^1_0 9 - 6e^{x \log 3} + e^{2x \log 3}\, dx$

$= \pi \left[ 9x - \frac{6}{\log 3}e^{x \log 3} + \frac{1}{2 \log 3}e^{2x \log 3} \right]^1_0$

$= \pi \left[\left(9 - \frac{6}{\log 3}e^{\log 3}+ \frac{1}{2 \log 3} e^{2 \log 3}\right)-\left(0-\frac{6}{\log 3} + \frac{1}{2 \log 3}\right) \right]$

Remember, $e^{\log x} = x,\, x > 0$

$= \pi \left(9 - \frac{18}{\log 3}+ \frac{9}{2 \log 3}+\frac{6}{\log 3} - \frac{1}{2 \log 3}\right)$

$= \pi \left(9 - \frac{12}{\log 3}+ \frac{8}{2 \log 3}\right)$

$= \pi \left(9 - \frac{12}{\log 3}+ \frac{4}{\log 3}\right)$

$= \pi \left(9 - \frac{8}{\log 3}\right) \text{u}^3$

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